很容易修复PHP(类)中的错误


Very easy to fix mistake in PHP (classes)

我刚开始处理类,但当我执行包含该类的页面时,我的脚本就停止了。此外,代码和函数本身根本不完整,只是添加了一些基本代码。

class doVersus
{
        private var $p1health;
        private var $p2health;
        private var $dmg;
        public var $longlog;
        function __construct($hp1m, $hp1c, $hp2m, $hp2c, $stg, $prev_log, $hp2n)
        {
                if($stg==0)
                {
                        $longlog = "The battle has begun..<br>".$_GET['user'];
                        //startBattle($hp1m, $hp1c, $hp2m, $hp2c, $stg, $prev_log, $hp2n);
                        echo $longlog;
                }else{
                        $longlog = $longlog."TURN: ".$stg."<br>";
                        //startBattle($hp1m, $hp1c, $hp2m, $hp2c, $stg, $prev_log, $hp2n);
                        echo $longLog;
        }
        private function startBattle($hp1m, $hp1c, $hp2m, $hp2c, $stg, $prev_log, $hp2n)
        {
                if($hp1c<0 || $hp2c<0)
                {
                        $longlog = $longlog."The battle has ended!<br>"
                        stopBattle($hp1c, $hp2c);
                }else{
                        echo $longlog;
                }
        }
        private function doDmg($a)
        {
                $longlog = $longlog." strikes ".$hp2n." with ".$dmg." normal damage";
        }
        private function stopBattle($a, $b)
        {
                echo $longlog." Battle has ended.";
        }
}

构造函数(__construct)的末尾缺少}

看起来构造函数的else没有关闭

您没有关闭构造函数中的else语句。

然后,在PHP5中,您不需要通过在声明中写入var来声明变量。